Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x-6y &= -9 \\ -x+4y &= 8\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $4y = x+8$ Divide both sides by $4$ to isolate $y$ $y = {\dfrac{1}{4}x + 2}$ Substitute this expression for $y$ in the first equation. $3x-6({\dfrac{1}{4}x + 2}) = -9$ $3x - \dfrac{3}{2}x - 12 = -9$ Simplify by combining terms, then solve for $x$ $\dfrac{3}{2}x - 12 = -9$ $\dfrac{3}{2}x = 3$ $x = 2$ Substitute $2$ for $x$ back into the top equation. $3( 2)-6y = -9$ $6-6y = -9$ $-6y = -15$ $y = \dfrac{5}{2}$ The solution is $\enspace x = 2, \enspace y = \dfrac{5}{2}$.